Question

If S is the circumcentre and O is the
orthocentre of triangle ABC, then OA+OB+OC =
(A) SO
(B) 2SO
(C) OS
(D) 2OS​

Answer 1

It has given that if S is the circumcentre and O is orthocentre of triangle ABC.

then we have to find OA + OB + OC

we know, centroid divides the distance from the circumcentre to orthocentre in the ratio of 2 : 1.

here,

OA + OB + OC = (A - O) + (B - O) + (C - O)

= (A + B + C) - 3O

= 3[(A + B + C)/3 - O]

= 3[G - O] [we know, centroid , G = (A + B + C)/3 ]

= 3GO ..........(1)

O----------(2)-------- G -----(1)----S

here, OG = 2GS

OS = OG + GS = GO + GO/2 = 3GO/2

⇒2OS = 3GO ..........(2)

from equations (1) and (2) we get,

OA + OB + OC = 2OS

therefore, option (D) is correct choice.

Answer 2

$\huge\mathcal{Answer:}$

↬Centroid divides the distance frm the circumcentre to the orthocentre in the ratio 2:1

$oa + ob + oc = (a - o) + (b - o) + (c - o)$

$= (a + b + c) - 30$

$= 3 \frac{(a + b + c)}{3 - o}$

$= 3(g - o)$

$centroid = g = \frac{(a + b + c)}{3}$

$3go - - - - - - - - - - - - (1)$

$o - - - (2) - - - g - - (1) - - s$

$og = 2gs$

$os = og + gs = go + \frac{go}{2} = \frac{3go}{2}$

$2os = 3go - - (2)$

$oa + ob + oc = 2os$